Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-z + 5}{-3z + 21} \div \dfrac{z^2 + 2z - 3}{z^2 - 4z - 21} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-z + 5}{-3z + 21} \times \dfrac{z^2 - 4z - 21}{z^2 + 2z - 3} $ First factor out any common factors. $n = \dfrac{-(z - 5)}{-3(z - 7)} \times \dfrac{z^2 - 4z - 21}{z^2 + 2z - 3} $ Then factor the quadratic expressions. $n = \dfrac {-(z - 5)} {-3(z - 7)} \times \dfrac {(z + 3)(z - 7)} {(z + 3)(z - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-(z - 5) \times (z + 3)(z - 7) } {-3(z - 7) \times (z + 3)(z - 1) } $ $n = \dfrac {-(z + 3)(z - 7)(z - 5)} {-3(z + 3)(z - 1)(z - 7)} $ Notice that $(z + 3)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-\cancel{(z + 3)}(z - 7)(z - 5)} {-3\cancel{(z + 3)}(z - 1)(z - 7)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $n = \dfrac {-\cancel{(z + 3)}\cancel{(z - 7)}(z - 5)} {-3\cancel{(z + 3)}(z - 1)\cancel{(z - 7)}} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $n = \dfrac {-(z - 5)} {-3(z - 1)} $ $ n = \dfrac{z - 5}{3(z - 1)}; z \neq -3; z \neq 7 $